Prove that the ratio of areas …

Given: Two triangles ABC and DEF such that ​{tex}\triangle {/tex}​ABC ​{tex} \sim {/tex}​ {tex}\triangle {/tex}DEF
To prove: {tex}\frac{{ar\left( {\vartriangle ABC} \right)}}{{ar\left( {\vartriangle DEF} \right)}} = \frac{{A{B^2}}}{{D{E^2}}} = \frac{{B{C^2}}}{{E{F^2}}} = \frac{{A{C^2}}}{{D{F^2}}}{/tex}
Construction: Draw AL ​{tex} \bot {/tex}​ BC and DM ​{tex} \bot {/tex}​ EF
Proof:- Since similar triangles are equiangular and their corresponding sides are proportional
​{tex}\therefore {/tex}​ ​{tex}\triangle {/tex}​ ABC ​{tex} \sim {/tex}​{tex}\triangle {/tex}DEF
​{tex}\Rightarrow {/tex}​ ​{tex}\angle{/tex}​ A = ​{tex}\angle{/tex}​ D, ​{tex}\angle{/tex}​ B = ​{tex}\angle{/tex}​ E, ​{tex}\angle{/tex}​ C ={tex}\angle{/tex}F
And {tex}\frac{{AB}}{{DE}} = \frac{{BC}}{{EF}} = \frac{{AC}}{{DF}}{/tex} .......(i)
In ​{tex}\triangle {/tex}​ ALB and ​{tex}\triangle {/tex}​ DMB,
​{tex}\angle{/tex}​ 1= ​{tex}\angle{/tex}​ 2 and ​{tex}\angle{/tex}​ B= ​{tex}\angle{/tex}​ E
​{tex}\Rightarrow {/tex}​ ​{tex}\triangle {/tex}​ ALB ​{tex} \sim {/tex}​ {tex}\triangle {/tex}DME [By AA similarity]
​{tex}\Rightarrow {/tex}​{tex}\frac{{AL}}{{DM}} = \frac{{AB}}{{DE}}{/tex} .....(ii)
From (i) and (ii), we get
{tex}\frac{{AB}}{{DE}} = \frac{{BC}}{{EF}} = \frac{{AC}}{{DF}} = \frac{{AL}}{{DM}}{/tex} ......(iii)
Now{tex}\frac{{area\left( {\vartriangle ABC} \right)}}{{area\left( {\vartriangle DEF} \right)}} = \frac{{\frac{1}{2}\left( {BC \times AL} \right)}}{{\frac{1}{2}\left( {BF \times DM} \right)}}{/tex}
{tex} \Rightarrow \frac{{area\left( {\vartriangle ABC} \right)}}{{area\left( {\vartriangle DEF} \right)}} = \frac{{BC}}{{EF}} \times \frac{{AL}}{{DM}}{/tex}
​​{tex} \Rightarrow \frac{{area\left( {\vartriangle ABC} \right)}}{{area\left( {\vartriangle DEF} \right)}} = \frac{{BC}}{{EF}} \times \frac{{BC}}{{EF}} = \frac{{B{C^2}}}{{E{F^2}}}{/tex}
Hence, {tex}\frac{{area\left( {\vartriangle ABC} \right)}}{{area\left( {\vartriangle DEF} \right)}} = \frac{{A{B^2}}}{{D{E^2}}} \times \frac{{B{C^2}}}{{E{F^2}}} = \frac{{A{C^2}}}{{D{F^2}}}{/tex}
Let the largest side of the largest triangle be x cm
Using above theorem,
{tex}\frac{{{x^2}}}{{{{27}^2}}} = \frac{{144}}{{81}} \Rightarrow \frac{x}{{27}} = \frac{{12}}{9}{/tex}
​{tex}\Rightarrow {/tex}​ x = 36 cm